A cannon and a target are 5.10Km apart and located at the same level. How soon will the shell launched with the initial velocity 240m/s reach the target in the absence of air drag?

Explanation

Here `v_(0) = 240 ms^(-1) ,R = 5.10 km = 5100m,g = 9.8 ms^(-2) , theta = ?`
From formula `R = (v_(0)^(2) sin 2theta)/(g)`
We have `sin 2 theta = (Rg)/(2)` , Putting values we get .
`sin 2 theta = (5100 xx 9.8)/(240 xx 240) = 0.8677 = (sqrt(3))/(2) = sin 60^(@) or sin 120^(@) rArr theta = 30^(@) or 60^(@)`
From formula ,` T = (2v_(0) sin theta)/(g)`
When `theta = 30^(@) , T_(1) = (2xx240 xx 0.5)/(9.8) = 24.5 sec` .
When `theta = 60^(@) , T_(2) = (2 xx 240 xx 0.867)/(9.8) = 42.41` sec.